Question: Add the following rational expressions. You may keep the denominator in its factored form. $\dfrac{5z^2}{60z-108}+\dfrac{40z^2}{15z^2-2z-45}=$
Solution: We can add two rational expressions whose denominators are equal by adding the numerators and keeping the denominator the same. When the denominators are not the same, we must manipulate them so that they become the same. More specifically, we should find the least common multiple of the two denominators. [What's that?] To do that, we first need to factor the denominators as much as possible: The first denominator $60z-108$ can be factored as ${(3)(4)}{(5z-9)}$ by factoring out a $12$. The second denominator $15z^2-2z-45$ can be factored as ${(5z-9)}{(3z+5)}$ using the method of grouping. We can see that: Both denominators share the factor ${(5z-9)}$. Only the first denominator has the factors ${(3)(4)}$. Only the second denominator has the factor ${(3z+5)}$. Therefore, the least common multiple is the product of all the above factors. Let's manipulate the expressions to have that denominator: $\begin{aligned}&\phantom{=}\dfrac{5z^2}{{(3)(4)}{(5z-9)}}+\dfrac{40z^2}{{(5z-9)}{(3z+5)}}\\\\\\\\ &=\dfrac{5z^2{(3z+5)}}{{(3)(4)}{(5z-9)}{(3z+5)}}+\dfrac{40z^2{(3)(4)}}{{(5z-9)}{(3z+5)}{(3)(4)}}\\\\\\\\ &=\dfrac{15z^3+25z^2}{12(5z-9)(3z+5)}+\dfrac{480z^2}{12(5z-9)(3z+5)}\end{aligned}$ Now that both denominators are the same, let's add! $\phantom{=}\dfrac{15z^3+25z^2}{12(5z-9)(3z+5)}+\dfrac{480z^2}{12(5z-9)(3z+5)}$ $\begin{aligned} &=\dfrac{(15z^3+25z^2)+(480z^2)}{12(5z-9)(3z+5)}&\text{Add numerators}\\\\\\\\ &=\dfrac{15z^3+25z^2+480z^2}{12(5z-9)(3z+5)}&\text{Distribute}\\\\\\\\ &=\dfrac{15z^3+505z^2}{12(5z-9)(3z+5)}&\text{Combine like terms}\end{aligned}$ In conclusion, $\dfrac{5z^2}{60z-108}+\dfrac{40z^2}{15z^2-2z-45}=\dfrac{15z^3+505z^2}{12(5z-9)(3z+5)}$